Synchronous Machines
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1 Synchronous Machines
2 Synchronous generators or alternators are used to convert mechanical power derived from steam, gas, or hydraulic-turbine to ac electric power Synchronous generators are the primary source of electrical energy we consume today Large ac power networks rely almost exclusively on synchronous generators
3 Synchronous Machines In a synchronous generator, a DC current is applied to the rotor winding producing a rotor magnetic field. The rotor is then turned by external means producing a rotating magnetic field, which induces a 3-phase voltage within the stator winding. Field windings are the windings producing the main magnetic field (rotor windings for synchronous machines); armature windings are the windings where the main voltage is induced (stator windings for synchronous machines
4 Basic parts of a synchronous generator: Rotor - dc excited winding Stator - 3-phase winding in which the ac emf is generated The manner in which the active parts of a synchronous machine are cooled determines its overall physical size and structure
5 Salient-pole synchronous machine Cylindrical or round-rotor synchronous machine
6 5.1 INTRODUCTION TO POLYPHASE SYNCHRONOUS MACHINES Two types: 1-Cylindirical rotor: High speed, fuel or gas fired power plants f e p 2 n 60 p 120 n To produce 50 Hz electricity p=2, n=3000 rpm p=4, n=1500 rpm 2-Salient-pole rotor: Low speed, hydroelectric power plants To produce 50 Hz electricity p=12, n=500 rpm p=24, n=250 rpm
7 Salient-Pole Synchronous Generator 1. Most hydraulic turbines have to turn at low speeds (between 50 and 300 r/min) 2. A large number of poles are required on the rotor d-axis N Non-uniform air-gap D 10 m q-axis S S Turbine Hydro (water) N Hydrogenerator
8 Stator
9 Cylindrical-Rotor Synchronous Generator Turbine D 1 m Steam L 10 m d-axis Stator winding High speed 3600 r/min 2-pole 1800 r/min 4-pole Direct-conductor cooling (using hydrogen or water as coolant) q-axis N Uniform air-gap Stator Rotor winding Rotor Rating up to 2000 MVA S Turbogenerator
10 Stator Cylindrical rotor
11 Slip rings Brush
12 The rotor of the generator is driven by a prime-mover A dc current is flowing in the rotor winding which produces a rotating magnetic field within the machine The rotating magnetic field induces a three-phase voltage in the stator winding of the generator
13 Electrical frequency produced is locked or synchronized to the mechanical speed of rotation of a synchronous generator: f e P n m 120 where f e = electrical frequency in Hz P = number of poles n m = mechanical speed of the rotor, in r/min
14 By the definition, synchronous generators produce electricity whose frequency is synchronized with the mechanical rotational speed. Steam turbines are most efficient when rotating at high speed; therefore, to generate 60 Hz, they are usually rotating at 3600 rpm and turn 2-pole generators. (7.11.1) Water turbines are most efficient when rotating at low speeds ( rpm); therefore, they usually turn generators with many poles.
15 The magnitude of internal generated voltage induced in a given stator is E 2N f K A where K is a constant representing the construction of the machine, is flux in it and is its rotation speed. C Since flux in the machine depends on the field current through it, the internal generated voltage is a function of the rotor field current. Magnetization curve (open-circuit characteristic) of a synchronous machine
16 The internal voltage E f produced in a machine is not usually the voltage that appears at the terminals of the generator. The only time E f is same as the output voltage of a phase is when there is no armature current flowing in the machine. There are a number of factors that cause the difference between E f and V : t o o o The distortion of the air-gap magnetic field by the current flowing in the stator, called the armature reaction The self-inductance of the armature coils. The resistance of the armature coils. The effect of salient-pole rotor shapes.
17 I a motor jx jx l R a generator I a E f E res V t Equivalent circuit of a cylindrical-rotor synchronous machine
18 5.2.4 EQUIVALENT CIRCUTS Motor: Generator: Vˆ a R a Iˆ a j X s Iˆ a Eˆ af Vˆ a Eˆ af R a Iˆ a j X s Iˆ a Synchronous Reactance
19 Often, armature reactance and self-inductance are combined into the synchronous reactance of the machine: X X X (7.18.1) S Therefore, the phase voltage is A V E A jx SI A RI A (7.18.2) The equivalent circuit of a 3-phase synchronous generator is shown. The adjustable resistor R adj controls the field current and, therefore, the rotor magnetic field.
20 The voltages and currents of the three phases are 120 o apart in angle, but otherwise the three phases are identical. + V t V L-L E f1 jx s R a + I a1 V L-L =3V t
21 Since the voltages in a synchronous generator are AC voltages, they are usually expressed as phasors. A vector plot of voltages and currents within one phase is called a phasor diagram. A phasor diagram of a synchronous generator with a unity power factor (resistive load) Lagging power factor (inductive load): a larger than for leading PF internal generated voltage E A is needed to form the same phase voltage. Leading power factor (capacitive load). For a given field current and magnitude of load current, the terminal voltage is lower for lagging loads and higher for leading loads.
22 A synchronous generator needs to be connected to a prime mover whose speed is reasonably constant (to ensure constant frequency of the generated voltage) for various loads. The applied mechanical power is partially converted to electricity P Pin appm 3E I cos conv ind m A A (7.22.1) (7.22.2) Where is the angle between E A and I A. The power-flow diagram of a synchronous generator.
23 The real output power of the synchronous generator is P 3V I cos 3V I cos out T L A (7.23.1) The reactive output power of the synchronous generator is Q 3V I sin 3V I sin out T L A (7.23.2) Recall that the power factor angle is the angle between V and I A and not the angle between V T and I L. In real synchronous machines of any size, the armature resistance R A << X S and, therefore, the armature resistance can be ignored. Thus, a simplified phasor diagram indicates that I A EA sin cos (7.23.3) X S
24 Then the real output power of the synchronous generator can be approximated as P out 3VE sin A (7.24.1) We observe that electrical losses are assumed to be zero since the resistance is neglected. Therefore: P conv X S P (7.24.2) Here is the torque angle of the machine the angle between V and E A. The maximum power can be supplied by the generator when = 90 0 : out P max 3VE A (7.24.3) X S
25 The maximum power specified by (7.24.3) is called the static stability limit of the generator. Normally, real generators do not approach this limit: full-load torque angles are usually between 15 0 and The induced torque is kb B kb B kb B sin (7.25.1) ind R S R net R net Notice that the torque angle is also the angle between the rotor magnetic field B R and the net magnetic field B net. Alternatively, the induced torque is ind 3VE sin A (7.25.2) X m S
26 The three quantities must be determined in order to describe the generator model: 1. The relationship between field current and flux (and therefore between the field current I F and the internal generated voltage E A ); 2. The synchronous reactance; 3. The armature resistance. We conduct first the open-circuit test on the synchronous generator: the generator is rotated at the rated speed, all the terminals are disconnected from loads, the field current is set to zero first. Next, the field current is increased in steps and the phase voltage (whish is equal to the internal generated voltage E A since the armature current is zero) is measured. Therefore, it is possible to plot the dependence of the internal generated voltage on the field current the open-circuit characteristic (OCC) of the generator.
27 Since the unsaturated core of the machine has a reluctance thousands times lower than the reluctance of the air-gap, the resulting flux increases linearly first. When the saturation is reached, the core reluctance greatly increases causing the flux to increase much slower with the increase of the mmf. We conduct next the short-circuit test on the synchronous generator: the generator is rotated at the rated speed, all the terminals are short-circuited through ammeters, the field current is set to zero first. Next, the field current is increased in steps and the armature current I A is measured as the field current is increased. The plot of armature current (or line current) vs. the field current is the short-circuit characteristic (SCC) of the generator.
28 The SCC is a straight line since, for the shortcircuited terminals, the magnitude of the armature current is I A R E A X 2 2 A S The equivalent generator s circuit during SC (7.28.1) The resulting phasor diagram Since B S almost cancels B R, the net field B net is very small. The magnetic fields during short-circuit test
29 An approximate method to determine the synchronous reactance X S at a given field current: 1. Get the internal generated voltage E A from the OCC at that field current. 2. Get the short-circuit current I A,SC at that field current from the SCC. 3. Find X S from X Since the internal machine impedance is S E I A A, SC E Z R X X since X R 2 2 A S A S S S A I A, SC (7.29.1) (7.29.2)
30 A 200 kva, 480-V, 60-Hz, 4-pole, Y-Connected synchronous generator with a rated field current of 5 A was tested and the following data was taken. a) from OC test terminal voltage = 540 V at rated field current b) from SC test line current = 300A at rated field current c) from Dc test DC voltage of 10 V applied to two terminals, a current of 25 A was measured. 1. Calculate the speed of rotation in r/min 2. Calculate the generated emf and saturated equivalent circuit parameters (armature resistance and synchronous reactance)
31 1. f e = electrical frequency = Pn m /120 f e = 60Hz + P = number of poles = 4 n m = mechanical speed of rotation in r/min. So, speed of rotation n m = 120 f e / P = (120 x 60)/4 = 1800 r/min E f j I a V t + 2. In open-circuit test, I a = 0 and E f =V t E f = 540/1.732 = V (as the machine is Y-connected) In short-circuit test, terminals are shorted, V t = 0 E f = I a Z s or Z s = E f /I a =311.8/300=1.04 ohm From the DC test, R a =V DC /(2I DC ) = 10/(2X25) = 0.2 ohm Synchronous reactance Z 2 2 s, sat Ra X s, sat X s 2, 2, sat Z s sat Ra
32 Example 7.2: A 480 V, 60 Hz, Y-connected six-pole synchronous generator has a per-phase synchronous reactance of 1.0. Its full-load armature current is 60 A at 0.8 PF lagging. Its friction and windage losses are 1.5 kw and core losses are 1.0 kw at 60 Hz at full load. Assume that the armature resistance (and, therefore, the I 2 R losses) can be ignored. The field current has been adjusted such that the no-load terminal voltage is 480 V. a. What is the speed of rotation of this generator? b. What is the terminal voltage of the generator if 1. It is loaded with the rated current at 0.8 PF lagging; 2. It is loaded with the rated current at 1.0 PF; 3. It is loaded with the rated current at 0.8 PF leading. c. What is the efficiency of this generator (ignoring the unknown electrical losses) when it is operating at the rated current and 0.8 PF lagging? d. How much shaft torque must be applied by the prime mover at the full load? how large is the induced countertorque? e. What is the voltage regulation of this generator at 0.8 PF lagging? at 1.0 PF? at 0.8 PF leading?
33 Since the generator is Y-connected, its phase voltage is V VT 3 277V At no load, the armature current I A = 0 and the internal generated voltage is E A = 277 V and it is constant since the field current was initially adjusted that way. a. The speed of rotation of a synchronous generator is which is m nm fe rpm P rad s 60 b.1. For the generator at the rated current and the 0.8 PF lagging, the phasor diagram is shown. The phase voltage is at 0 0, the magnitude of E A is 277 V,
34 and that jx I S A j Two unknown quantities are the magnitude of V and the angle of E A. From the phasor diagram: sin cos A S A S A E V X I X I Then: 2 V E X I cos X I sin 236.8V 2 A S A S A Since the generator is Y-connected, V 3V 410V T
35 b.2. For the generator at the rated current and the 1.0 PF, the phasor diagram is shown. Then: 2 V E X I cos X I sin V 2 A S A S A and V 3V V T b.3. For the generator at the rated current and the 0.8 PF leading, the phasor diagram is shown. Then: 2 V E X I cos X I sin 308.8V 2 A S A S A and V 3V 535V T
36 c. The output power of the generator at 60 A and 0.8 PF lagging is P 3V I cos kw out The mechanical input power is given by A P P P P P kw in out elec loss core loss mech loss The efficiency is Pout % 100% 93.2% P 36.6 in d. The input torque of the generator is app Pin N - m m
37 The induced countertorque of the generator is app e. The voltage regulation of the generator is Lagging PF: Unity PF: Lagging PF: Pconv N - m m VR 100% 17.1% VR 100% 2.6% VR 100% 10.3% 535
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